I need help with math I tried my best but couldn’t get it

Step-by-step explanation:

1. Convert to slope intercept form (y=mx+b)

2. Find slope (the 'm' in mx)

3. Find slope of perpendicular line

So for the first question:

Convert to slope Intercept form if necessary:

3x+4y=12

4y=12-3x<<Subtract by 3x

y=3-3/4x<<Divide by 4

Find slope

y=3-3/4x

3/4 is the slope

Find slope of perpendicular line

To find the slope of a perpendicular line, you just have to take the reciprocal of the original slope and take the opposite sign.

Here is what I mean:

3/4<<original slope

4/3<<reciprocal

-4/3<<opposite

So for the first question -4/3 is the answer.

Now you can do the rest! Hope this helps!

(If you have more questions feel free to comment to let me know!)

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.